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Difficulty: MediumRational and Radical Expressions and Equations

Solve the equation 2x+7−x=−4\sqrt{2x + 7} - x = -4 for xx. What is the value of the real solution?

Answer: 9

Answer

The only real solution to the equation is 9.
Isolating the radical yields 2x+7=x−4\sqrt{2x + 7} = x - 4. Squaring both sides results in the quadratic equation 2x+7=x2−8x+162x + 7 = x^2 - 8x + 16, which simplifies to x2−10x+9=0x^2 - 10x + 9 = 0. Factoring gives (x−9)(x−1)=0(x - 9)(x - 1) = 0, yielding potential solutions of 9 and 1. Checking these solutions in the original equation reveals that 9 is valid, while 1 is extraneous. Therefore, the correct real solution is 9.

Step-by-Step Solution

1
Isolate the radical on one side of the equation.
2x+7=x−4\sqrt{2x + 7} = x - 4
Before squaring both sides, the radical term must be isolated to simplify the algebraic manipulation.
2
Square both sides of the equation.
2x+7=(x−4)22x + 7 = (x - 4)^2
Squaring both sides eliminates the square root.
3
Expand the squared binomial.
2x+7=x2−8x+162x + 7 = x^2 - 8x + 16
Applying the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 is necessary to write the equation in polynomial form.
4
Set the quadratic equation to zero.
x2−10x+9=0x^2 - 10x + 9 = 0
Subtracting 2x2x and 77 from both sides allows us to solve the quadratic equation.
5
Factor the quadratic equation.
(x−9)(x−1)=0(x - 9)(x - 1) = 0, giving potential solutions x=9x = 9 or x=1x = 1.
Finding the roots of the quadratic equation provides the candidate solutions.
6
Verify candidates in the original equation.
The solution x=9x = 9 is valid, while x=1x = 1 is extraneous.
Squaring both sides can introduce extraneous solutions, so each candidate must be checked in the original equation.

Key Concept

Solving radical equations and verifying for extraneous solutions
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Topics

Rational and Radical Expressions and Equations

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